Question: Simplify and expand the following expression: $ \dfrac{t + 1}{t - 8}-\dfrac{4t - 3}{t - 10} $
Solution: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(t - 8)(t - 10)$ Multiply the first term by $\dfrac{t - 10}{t - 10}$ $ \begin{align*} \dfrac{t + 1}{t - 8} \times \dfrac{t - 10}{t - 10} & = \dfrac{(t + 1)(t - 10)}{(t - 8)(t - 10)} \\ & = \dfrac{t^2 - 9t - 10}{(t - 8)(t - 10)}\end{align*} $ Multiply the second term by $\dfrac{t - 8}{t - 8}$ $ \begin{align*} \dfrac{4t - 3}{t - 10} \times \dfrac{t - 8}{t - 8} & = \dfrac{(4t - 3)(t - 8)}{(t - 10)(t - 8)} \\ & = \dfrac{4t^2 - 35t + 24}{(t - 10)(t - 8)}\end{align*} $ Now we have: $ = \dfrac{t^2 - 9t - 10}{(t - 8)(t - 10)} - \dfrac{4t^2 - 35t + 24}{(t - 10)(t - 8)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{t^2 - 9t - 10 - (4t^2 - 35t + 24)}{(t - 8)(t - 10)} $ $ = \dfrac{t^2 - 9t - 10 - 4t^2 + 35t - 24}{(t - 8)(t - 10)} $ $ = \dfrac{-3t^2 + 26t - 34}{(t - 8)(t - 10)}$ Expand the denominator: $ = \dfrac{-3t^2 + 26t - 34}{t^2 - 18t + 80}$